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180 lines
3.2 KiB
Markdown
180 lines
3.2 KiB
Markdown
Greedy
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------
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$k$ negations
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-------------
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> Given A[N]
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> can have -ve can have 0s
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> k negations
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> can choose same element multiple times
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> find the maximum sum that you can achieve
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always find the smallest and negate it
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Sort: $O(n \log n + k)$
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Min Heap: $O(n+k \log n)$
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Note: 1 2 will have -1 2 or 1 2 as the final answer
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can keep toggling the smallest element
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-- --
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What is Greedy
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--------------
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Greedy - when optimal solution to a local problem is also the optimal solution to the global problem
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That is, if you can act greedily at every step, and still reach the optimal.
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If you don't have to think long term to achieve the optimal
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-- --
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Where will greedy not work?
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---------------------------
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Local solution is not contributing to the global solution
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Whenever greedy works, it will be based on observations and intuition
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But have to be careful because can be subtle-non-greedy case
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-- --
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Max Sum Subset
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--------------
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> Given A[N] with -ves
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> Find the subset of size k with maximum sum
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>
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Sort and take top k values
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$O(n \log n)$
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-- --
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Amazon - Min Subset Product
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----------------------------
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> Given A[N]
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> min possible subset product
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>
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Examples
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> -1 -1 -2 4 5 = -24
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> -1 0 2 = -2
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> 0 1 = 0
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>
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1. If no -ve, then choose min $\geq 0$
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2. If k -ve, then choose all $> 0$. don't choose 0
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- if odd -ve, choose all
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- if even -ve, choose the smallest odd -ves
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$O(n)$
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-- --
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Activity Selection
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------------------
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Multiple companies. Very important
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> 
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> max no of activity that can be performed without overlapps
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**Incorrect:** min sized?
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```
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============= ===============
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========
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```
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**Incorrect:** start time
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**Correct:** least overlaps?
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**but easier:** first one to end
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sort by end time and execute
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**Proof:**
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Assume there exists a solution that doesn't include the min end time job.
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Add this job. Either it overlaps with 1, or with 0.
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Optimal in both cases
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So, for every optimal solution without the min end job, we also have an optimal solution with the min end job.
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So, optimal to always choose the min end job
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-- --
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Flights
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-------
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> start, end time of flights
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> num of platforms
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>
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+1 -1, prefix sum max
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-- --
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Job Scheduling + Profit
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-----------------------
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> day deadlines for jobs
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> can perform only 1 job per day
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delay the job
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$O(n \log n + n^2)$
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$n \log n$ to sort by profit
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$n^2$ because start at end time and move left to find an empty slot
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Can reduce to n log n using BBST
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we want the highest number in the set which is less than x
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we also want to remove any given number from set
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set - upperbound - google
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-- --
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Rats and Holes
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--------------
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> Rats
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> Holes
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> equal count
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> locations given for each
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> ech rat runs together
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> each must reach a hole
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> max time so that all reach hole
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>
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sort, all go left or all go right
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cant cross because that will increase time
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```
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------- ----------
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vs
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----------
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---------------------------
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``` |